A) 2.8 atm
B) 2atm
C) 3.2 atm
D) 4 atm
Correct Answer: B
Solution :
\[{{N}_{2}}+3{{H}_{2}}\xrightarrow[{}]{{}}2N{{H}_{3}}\] Initial moles \[2~~~~~~~~~~5~~~~~~~~~~~~~0\] reacted (50% \[{{N}_{2}}\]) \[~1~~~~~~~~~3~~~~~~~~~~~~~~2\] Left moles. \[1~~~~~~~~~2~~~~~~~~~~~~~~2\] Total number of moles of\[{{N}_{2}}{{H}_{2}}\]and\[N{{H}_{3}}\]left \[=1+2+2=5\] Initial moles = 7 pressure at initials =7 atm pressure at equilibrium =5 atm Partial pressure of\[N{{H}_{3}}=\frac{2}{5}\times 5=2\,atm\]You need to login to perform this action.
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